hdu 4497 GCD and LCM

標籤：des   style   blog   http   color   io   os   ar   java   

GCD and LCM

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1092    Accepted Submission(s): 512

Problem DescriptionGiven two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and lcm(x, y, z) = L? 
Note, gcd(x, y, z) means the greatest common divisor of x, y and z, while lcm(x, y, z) means the least common multiple of x, y and z. 
Note 2, (1, 2, 3) and (1, 3, 2) are two different solutions. 

 

InputFirst line comes an integer T (T <= 12), telling the number of test cases. 
The next T lines, each contains two positive 32-bit signed integers, G and L. 
It’s guaranteed that each answer will fit in a 32-bit signed integer. 

 

OutputFor each test case, print one line with the number of solutions satisfying the conditions above. 

 

Sample Input2 6 72 7 33  

 

Sample Output72 0 

 

Source2013 ACM-ICPC吉林通化全國邀請賽——題目重現 

題意:求滿足 gcd(x,y,z) = g , lcm(x，y,z) = l  的 組數

首先 l%g != 0 肯定是不行的，

我們對 l 和 g 經行素數分解

g = p1^a1 * p2 ^ a2 * p3 ^ a3 ...

l = p1^b1 * p2 ^ b2 * p3 ^ b3 ...

因為 l %g == 0 ;所以 ai <= bi 對於p1 素數，對於 x,y,z ，需要有一個數的素數分解 中 p1的次數為 ai ；一個數的素數分解 中 p1的次數為 bi 如果 ai == bi ，那麼 沒有選擇ai < bi 的選擇就是 C(1,3)*C(1,2)*(bi-a1+1) 公式就是 6*(bi-ai+1) 意思就是，三個裡面選一個取 ai，然後兩個裡面選一個選 bi ，第三個的任意選

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<vector>#include<set>#include<stack>#include<map>#include<ctime>#include<bitset>#define LL long long#define mod 1000000007#define maxn 110#define pi acos(-1.0)#define eps 1e-8#define INF 0x3f3f3f3fusing namespace std;bool check(LL i ){    for(int j = 2 ; j*j <= i ;j++)if(i%j == 0)        return false;    return true;}int main(){    int j,i,l,g;    int T,ans1,u,v;    LL ans;    cin >> T ;    while(T--)    {        scanf("%d%d",&g,&l ) ;        if(l%g !=0)puts("0") ;        else        {            l /= g ;            ans=1;            for( LL i = 2 ; i*i <= l ;i++)if(l%i==0&&check(i))            {                v = 0;                while(l%i==0)                {                    v++;                    l /= i ;                }                ans *= v*6 ;            }            if(l !=1) ans *= 6 ;            printf("%I64d\n",ans);        }    }    return 0 ;}

View Code

 

  

hdu 4497 GCD and LCM